LeetCode 題解(254) : Meeting Rooms

題目：

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (s_i < e_i), determine if a person could attend all meetings.

For example,
Given [[0, 30],[5, 10],[15, 20]],
return false.

題解：

按start排序，然後比較前一個的end和後一個的start。

C++版：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    bool canAttendMeetings(vector<Interval>& intervals) {
        sort(intervals.begin(), intervals.end(), compare);
        for(int i = 1; i < intervals.size(); i++) {
            if(intervals[i - 1].end > intervals[i].start)
                return false;
        }
        return true;
    }
    
    static bool compare(Interval a, Interval b) {
        return a.start < b.start;
    }
};

Java版：

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
import java.util.Arrays; 
 
public class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals, new IntervalComparator());
        for(int i = 1; i < intervals.length; i++) {
            if(intervals[i - 1].end > intervals[i].start)
                return false;
        }
        return true;
    }
}

class IntervalComparator implements Comparator<Interval> {
    public int compare(Interval a, Interval b) {
        return a.start - b.start;
    }
}

Python版：

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def canAttendMeetings(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: bool
        """
        intervals.sort(lambda a, b : a.start - b.start)
        for i in range(1, len(intervals)):
            if intervals[i - 1].end > intervals[i].start:
                return False
        return True